Problem: Consider the polar curve $r=1-\sin(\theta)$ for $0\le \theta < 2\pi$. At which values of $\theta$ does the graph of $r$ have a horizontal tangent line? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{\pi}{2}$ or $\dfrac{3\pi}{2}$ only (Choice B) B $\dfrac{\pi}{6}$ or $\dfrac{5\pi}{6}$ only (Choice C) C $\dfrac{\pi}{6}$, $\dfrac{5\pi}{6}$, or $\dfrac{3\pi}{2}$ only (Choice D) D $\dfrac{\pi}{6}$, $\dfrac{\pi}{2}$, $\dfrac{5\pi}{6}$, or $\dfrac{3\pi}{2}$
Solution: A horizontal line has a slope of $0$. The slope of the tangent line at a point is equal to $\dfrac{dy}{dx}$ at that point. In the case of polar curves, we can use the relationship: $\dfrac{dy}{dx}=\dfrac{\left( \dfrac{dy}{d\theta} \right)}{\left( \dfrac{dx}{d\theta} \right)}$ Once we have an expression for the slope of the tangent line, we can look for the $\theta$ -values that make $\dfrac{dy}{d\theta}=0$ but don't make $\dfrac{dx}{d\theta}=0$ (because then $\dfrac{dy}{dx}$ will be undefined). For a polar curve, $x={r}\cos(\theta)$ and $y={r}\sin(\theta)$. Therefore, in our problem we have: $\begin{aligned} x&={(1-\sin(\theta))}\cos(\theta) \\\\ &=\cos(\theta)-\sin(\theta)\cos(\theta) \\\\ &=\cos(\theta)-\dfrac{1}{2}\sin(2\theta) \\\\ \\\\ y&={(1-\sin(\theta))} \sin(\theta) \\\\ &=\sin(\theta)-\sin^2(\theta) \end{aligned}$ Let's find $\dfrac{dy}{d\theta}$ and $\dfrac{dx}{d\theta}$. $\begin{aligned} y(\theta)&=\sin(\theta)-\sin^2(\theta) \\\\ \dfrac{dy}{d\theta}&=\cos(\theta)-2\sin(\theta)\cos(\theta) \\\\ &=\cos(\theta)-2\sin(\theta)\cos(\theta) \\\\ \\\\ x(\theta)&=\cos(\theta)-\dfrac{1}{2}\sin(2\theta) \\\\ \dfrac{dx}{d\theta}&=-\sin(\theta)-\dfrac{1}{2}\cos(2\theta)(2) \\\\ &=-\sin(\theta)-\cos(2\theta) \end{aligned}$ Now let's solve $\dfrac{dy}{d\theta}=0$ on the interval $0\le \theta < 2\pi$. $\begin{aligned} \dfrac{dy}{d\theta}&=0 \\\\ \cos(\theta)-2\sin(\theta)\cos(\theta)&=0 \\\\ \cos(\theta)(1-2\sin(\theta))&=0 \\\\ \cos(\theta)=0&\text{ or }\sin(\theta)=\dfrac{1}{2} \end{aligned}$ Within our interval, our possible solutions are $\theta=\dfrac{\pi}{2}$, $\theta=\dfrac{3\pi}{2}$, $\theta=\dfrac{\pi}{6}$, or $\theta=\dfrac{5\pi}{6}$. Finally, we evaluate $\dfrac{dx}{d\theta}$ for our four possible values of $\theta$ and require that $\dfrac{dx}{d\theta}\ne0$. $\begin{aligned} \left.\dfrac{dx}{d\theta} \right| _{{\theta =\tfrac{\pi }{2}}}&=-\sin\left({\dfrac{\pi}{2}}\right)-\cos\left(2\left({\dfrac{\pi}{2}}\right)\right) \\\\ &=-(1)-\cos(\pi) \\\\ &=-1-(-1) \\\\ &=0 \\\\ \\\\ \left.\dfrac{dx}{d\theta} \right| _{{\theta =\tfrac{3\pi }{2}}}&=-\sin\left({\dfrac{3\pi}{2}}\right)-\cos\left(2\left({\dfrac{3\pi}{2}}\right)\right) \\\\ &=-(-1)-\cos(3\pi) \\\\ &=1-(-1) \\\\ &=2 \\\\ \\\\ \left.\dfrac{dx}{d\theta} \right| _{\theta =\tfrac{\pi }{6}}}&=-\sin\left(\dfrac{\pi}{6}}\right)-\cos\left(2\left(\dfrac{\pi}{6}}\right)\right) \\\\ &=-\left(\dfrac{1}{2}\right)-\cos\left(\dfrac{\pi}{3}\right) \\\\ &=-\dfrac{1}{2}-\dfrac{1}{2} \\\\ &=-1 \\\\ \\\\ \left.\dfrac{dx}{d\theta} \right| _{{\theta =\tfrac{5\pi }{6}}}&=-\sin\left({\dfrac{5\pi}{6}}\right)-\cos\left(2\left({\dfrac{5\pi}{6}}\right)\right) \\\\ &=-\left(\dfrac{1}{2}\right)-\cos\left(\dfrac{5\pi}{3}\right) \\\\ &=-\dfrac{1}{2}-\dfrac{1}{2} \\\\ &=-1 \end{aligned}$ The graph of $r$ has a horizontal tangent at $\theta=\dfrac{\pi}{6}$ or $\theta=\dfrac{5\pi}{6}$ or $\theta=\dfrac{3\pi}{2}$. Note : It turns out that the curve is not differentiable at $\theta=\dfrac{\pi}{2}$, where the graph comes to a "point," as seen below. The graphs of the tangents are shown. ${0.5}$ ${1}$ ${1.5}$ ${2}$ ${0}$ ${\frac{1}{6}\pi}$ ${\frac{1}{3}\pi}$ ${\frac{1}{2}\pi}$ ${\frac{2}{3}\pi}$ ${\frac{5}{6}\pi}$ ${\pi}$ ${\frac{7}{6}\pi}$ ${\frac{4}{3}\pi}$ ${\frac{3}{2}\pi}$ ${\frac{5}{3}\pi}$ ${\frac{11}{6}\pi}$